Guided Media are those media that provide a conduit from one device to another. Guided

Transmission Media uses a “cabling” system that guides the data signals along a specific path while
Unguided Transmission Media consists of a means for the data signals to travel but nothing to guide
them along a specific path. It passes through a vacuum; it is independent of a physical pathway.

o File Transfer protocol is similar in operation to XMODEM, with sender waiting for an
NAK before it starts TX

o Kermit allows the transmission of control characters as Text

Question#2

What is Frequency division multiplexing ?…… .Marks (5)

Answer: (Page 149)

Frequency division multiplexing (FDM)

o An analog technique that can be applied when BW of the link is greater than the combined
BW of the signals to be TX

o Signals generated by each sending device modulate difference carrier frequencies o These modulated signals are then combined into a single Composite signal that
can be transported by the link

o Carrier frequencies are separated by enough BW to accommodate the modulated signal o These BW ranges are the channels through which the various signals travel

Question#4

What is stop and wait ARQ in error control ?….Marks (3)

Answer: Page 197

Stop-and-Wait is an extended form of flow control to include retransmission of data in case of Lost or Damaged frames.

There are four main features added in it.

1. Sending device keeps a copy of the last frame transmitted until it receives the ACK for that
frame.

2. Both data and ACK frames are numbered 0 and 1 alternately.

3. A data 0 frame is acknowledged by a ACK 1 frame indicating that the receiver has received data
0 and is now expecting data 1 .

4. For retransmission to work, 4 features are added to the basic flow control mechanism.

Question#5

What is Interleaving ?… ..Marks (3)

Answer: Page 153

Synchronous TDM is considered as a very fast rotating switch. When this switch opens in front of a
device, the device has the opportunity to send a specific amount of data on to the path.
The switch moves from device to device at a constant rate and in a fixed order. This process is called
INTERLEAVING. Interleaving can be done by BITS, BYTES or by any other DATA UNIT

Question#6

What is DSU in terms of digital services?… …Marks (3)

Answer: Page 163

DSU (Digital service unit) changes the rate of digital data created by the subscriber’s device to 56 Kbps and encodes it in the format used by service provider. It used in dialing process and is more expensive than MODEM. But it has better speed, better quality and less susceptibility to noise.

Question#7

Which architecture of Ethernet developed by ITU_T and

ANSI?…….. .. Marks (2)

Answer: 236

FDDI (Fiber Distributed Data Interface)architecture of Ethernet developed by ITU_T and ANSI.

Question#8

What is a spike in noise term?… …Marks (2)

Answer:143

Spike is a signal with high energy in a very short period of time that comes from power lines, lightening etc,

Question#9

Answer: Page 172

What is even parity generator in VRC error detection mechanism?… ..Marks (2)

Even parity generator counts the 1’s and appends the parity bit (1) to the end.

Addressing is required in Poll / Select method as it is a not point-to-point configuration, For the primary device in a multipoint topology to be able to identify and communicate with a specific secondary
device, there must be some addressing, while ENQ/ACK method is a point-to-point method and for point-to-point configuration, there is no need for addressing.

Still each node must be connected to a Hub , so Cabling is still much more

What are the two major classes of synchronous protocols at data link layer?2

Answer: (Page 206)

Character – Oriented Protocols
Bit – Oriented Protocols

Whether Hamming code is the technique used for error detection or error correction?2

Answer: (Page 181)

Hamming code is the technique used for error correction

Define Multiplexing? What is its advantage?2

Answer: (Page 147)

Set of techniques that allows the simultaneous transmission of multiple signals across a single data link

It allows multiple users to share total capacity of a Transmission Medium.

What is the purpose of dual ring?2

Answer: (Page 34)

Unidirectional traffic movement is overcome by dual ring technology.

Which modem was first developed commercially in 1970?2

Answer: (Page 114)

Bell modems

Write any two functions of physical layer?2

Answer: (Page 45)

It defines characteristics of Interface between device and transmission Medium It also defines the type of transmission medium

Physical Layer is also concerned with Line Configuration

FINAL TERM EXAMINATION

2011

Which one has more overhead, a repeater or a bridge? Explain your answer. [3]

Answer:

A bridge has more overhead than a repeater. A bridge processes the packet at two layers; a repeater processes a frame at only one layer. A bridge needs to search a table and find the forwarding port as well as to regenerate the signal; a repeater only regenerates the signal. In other words, a bridge is also a repeater (and more); a repeater is not a bridge

Define high frequency [HF] and super high frequency [SHF], which devices uses these frequencies [3]

Answer: Page 135 and 136

High frequency.

HF uses ionospheric propagation. These frequencies move into the ionosphere where the density difference reflects them back on earth.

It is used for Citizen’s Band Radio, International Broadcasting, Military Communication, Telephone, Telegraph and Fax

Super high frequency.

SHF waves are TX using mostly line-of-sight and some Space propagation.

It is used for Terrestrial and Satellite Microwave and Radar Communication devices.

Write all steps of checksum method. [3]

Answer: (Page 179)

o The sender subdivides data units into equal segments of ‘n’ bits(16 bits) o These segments are added together using one’s complement

o The total (sum) is then complemented and appended to the end of the original data unit as redundancy bits called CHECKSUM

o The extended data unit is transmitted across the network

o The receiver subdivides data unit as above and adds all segments together and complement the result o If the intended data unit is intact, total value found by adding the data segments and the checksum field should be zero o If the result is not zero, the packet contains an error & the receiver rejects it

Differentiate internet and the internet? [3]

Answer: (Page 240)

INTERNET

o An internet is a generic term used to mean an interconnection of individual networks o To create an internet, we need networking devices called routers and gateways
o An internet is different from the Internet

o Internet is the name of a specific worldwide network

What is the differences in between bit oriented and character oriented protocols [5]

Answer: (Page 206)

Character – Oriented Protocols

o Also called Byte- Oriented Protocol

o These protocols interpret a transmission frame or packet as a succession of characters,
each usually composed of one byte

o All control information is in the form of an existing character encoding system

Bit – Oriented Protocols

o Character -Oriented Protocols are not as efficient as bit – oriented protocols and are seldom used o They are easy to comprehend and employ the same logic as bit-oriented protocols
o Their study will provide the basis for studying the other data link layer protocols
o IBN’s BSC is the best known character oriented protocol

FINAL TERM EXAMINATION

2011

Question#1

What is the formula to calculate the number of redundancy bits required to correct a bit error in a given number of data bits? [2]

Broadcast: transmitting a packet that will be received by every device on the network Unicast: the sending of information packets to a single destination
Multicast: delivery of information to a group of destinations.
Question#5

T lines are designed for Digital data how they can be used for Analog Transmission?

Answer: (Page 166)

o DS-1 requires 8 Kbps of overhead

o To understand this overhead, let’s examine format of a 24-voice channel frame

o Frame used on T-1 line is usually 193 bits divided into 24 slots of 8 bits each + 1 bit for synchronization (24*8+1=193)

o 24 segments are interleaved in one frame

o If a T-1 carries 8000 frames, the data rate is 1.544 Mbps (193 * 8000=1.544 Mbps) which is capacity of the line

Question#6

What are the three types of Guided Media?

Answer: (Page 120)

1. Coaxial cable

2. Twisted-pair cable

3. Fiber optic cable.

Question#7

Why do we need Inverse Multiplexing? [5]

Answer: (Page 159)

 An organization wants to send data, voice and video each of which requires a different data rate  To send voice it needs 64Kbps,

 To send data, it needs 128 Kbps link

 To send video it may need 1.544 Mbps link

 It can lease a 1.544 Mbps line from a common carrier and only use it fully for sometime  Or it can lease several separate channels of lower data rates

 Voice can be sent over any of these channels

 Data & Video can be broken into smaller portions using Inverse Multiplexing and TX

Question#8

Describe method of checksum briefly?

Answer: (Page 180)

o The sender subdivides data units into equal segments of ‘n’ bits(16 bits) o These segments are added together using one’s complement

o The total (sum) is then complemented and appended to the end of the original data unit as redundancy bits called CHECKSUM

o The extended data unit is transmitted across the network

o The receiver subdivides data unit as above and adds all segments together and complement the result o If the intended data unit is intact, total value found by adding the data segments and the checksum field should be zero o If the result is not zero, the packet contains an error & the receiver rejects it

Question#9

Explain Asynchronous Time Division Multiplexing in detail? Also discuss its advantages over synchronous TDM?

Answer:

Asynchronous time-division multiplexing (ATDM) is a method of sending information that resembles
normal TDM, except that time slots are allocated as needed dynamically rather than pre-assigned to
specific transmitters. ATDM is more intelligent and has better bandwidth efficiency than TDM.
asynchronous time-division multiplexing comprising receive circuits (CRl/i) supplying cells received
via input links, transmit circuits (CTl/j) transmitting retransmitted cells on output links, a buffer
memory (MT) storing the received cells and delivering the cells to be retransmitted and a buffer
memory addressing device (SMT) including a write address source (SAE) and a read address source
(fsl/j).

Advantages asynchronous TDM:

In asynchronous TDM, the timeslots are not fixed. They are assigned dynamically as needed.

The objective would be to switch from one user to another user whenever the one user is idle, and to
asynchronously time multiplex the data. With such an arrangement, each user would be granted access
to the channel only when he has a message to transmit. This is known as an Asynchronous Time
Division Multiplexing System (ATDM). A segment of a typical ATDM data stream is shown in Figure

2. The crucial attributes of such a multiplexing technique are:

1. An address is required for each transmitted message, and

2. Buffering is required to handle the random message arrivals.

FINAL TERM EXAMINATION

2011

Question No: 31 ( Marks: 2 )

What is hybrid topology?

Answer:

Hybrid topology is a kind of topology, In which Several topologies combined in a larger topology

Question No: 32 ( Marks: 2 )

What is combined station of HDLC?

Answer: (Page 211)

A combined station is one of a set of connected peer devices programmed to behave either as a primary or as a secondary depending on the nature and the direction of the transmission.

Question No: 33 ( Marks: 2 )

What kind of error is undetectable by the checksum?

Answer: (Page 180)

Error is invisible if a bit inversion is balanced by an opposite bit inversion in the corresponding digit of another segment

Question No: 34 ( Marks: 2 )

What’s the name of the telephone service in which there is no need of dialing?

Answer: (Page 161)

In Analog Leased Service there is no need of dialing

Question No: 35 ( Marks: 3 )

How Bit Rate & Baud rate are related?

Answer: (Page 85)

Bit rate equals the baud rate times the no. of bits represented by each signal unit.

The baud rate equals the bit rate divided by the no. of bits represented by each signal shift. Bit rate is always greater than or equal to Baud rate

Question No: 36 ( Marks: 3 )

Following abbreviations stands for what?

Answer: (Page244)

1. ARP: Address Resolution Protocol

2. RARP: Reverse Address Resolution Protocol

3. ICMP: Internet Control Message Protocol

Question No: 37 ( Marks: 3 )

Differentiate between Polling and Selecting.

Answer: (Page 189)

If the primary wants to receive data, it asks the second-arise if they have anything to send, This is called Polling

If the primary wants to send data, it tells the target secondary to get ready to receive, This function is called Selecting.

Question No: 38 ( Marks: 3 )

T lines are designed for Digital data how they can be used for Analog Transmission?

Answer: repeat

Question No: 39 ( Marks: 5 )

What is the difference between character oriented and bit oriented protocols?

Answer: repeat

Question No: 40 ( Marks: 5 )

Write down the function of Primary-Secondary communication in line discipline.

Answer: (Page 189)

1. Poll method works with topologies where one device is designed as a Primary station
and the other devices are Secondary stations

3. It is up to the primary to determine which device is allowed to use the channel at a
given time.

4. The primary therefore is always the initiator of the a session

5. Whenever a multipoint link consists of a primary device and multiple secondary devices
using a single TX line , all exchanges must be made through the primary device even
when the ultimate destination is a secondary device

FINAL TERM EXAMINATION

2010

Question No: 31 ( Marks: 2 )

What kind of error is undetectable by the checksum? [2]

Answer: Page 180 (repeated)

Question No: 32 ( Marks: 2 )

What are properties of signals?

Answer: (Page 17)

Capable of being propagated over TX. Medium ,Interpretable as data at the Receiver

A router is a device that extracts the destination of a packet it receives, selects the best path to that destination, and forwards data packets to the next device along this path. They connect networks together; a LAN to a WAN for example, to access the Internet.

Question No: 35 ( Marks: 3 )

Why we need a Null Modem?

Answer: (Page 106)

A null modem provide DTA -DTE interface w/o DCEs

Question No: 36 ( Marks: 3 )

Count LRC for the following bits?

10011010 10100101 1101 0110

Question No: 37 ( Marks: 3 )

What are the categories of multiplexing?

Answer: Page 148

There are three catogaries of multiplxing
FDM

TDM

Have a two other catagories
Synchronous and asyrounce
WDM

Question No: 38 ( Marks: 3 )

What are the three purposes of control frames at data link layer?

Answer: (Page 209)

Control frames serve 3 purposes:

Establishing Connections

Maintaining Flow and Error Control during Data Transmission Terminating Connection

Question No: 39 ( Marks: 5 )

Compare line discipline methods ENQ/ACK and Poll/ Select?

Answer: (Page 188-189)

=>ENQ/ACK coordinates which device may start a transmission and whether or not the intended recipient is ready and enabled.

=> Using ENQ/ACK, a session can be initiated by either station on a link as long as both are of equal
rank.

=> In both half-duplex and full-duplex transmission, the initiating device establishes the session.

=> In half duplex, the initiator then sends its data while the responder waits. The responder may take over the link when the initiator is finished or has requested a response.

=> In full duplex, both devices can transmit simultaneously once the session has been established.

POLL/SELECT:

=> The poll/select method of line discipline works with topologies where one device is designated as a primary station and the other devices are secondary stations.

2. The total (sum) is then complemented and appended to the end of the original data unit
as redundancy bits called CHECKSUM.

3. The extended data unit is transmitted across the network.

4. The receiver subdivides data unit and adds all segments together and complement the
result.

5. If the intended data unit is intact, total value found by adding the data segments and the
checksum field should be zero.

6. If the result is not zero, the packet contains an error & the receiver rejects it.

Question No: 36 ( Marks: 3 )

Which one has more overhead, a repeater or a bridge? Explain your answer. [3]

Answer:

A bridge has more overhead than a repeater. A bridge processes the packet at two layers; a repeater processes a frame at only one layer. A bridge needs to search a table and find the forwarding port as well as to regenerate the signal; a repeater only regenerates the signal. In other words, a bridge is also a repeater (and more); a repeater is not a bridge.

Question No: 37 ( Marks: 3 )

Write down disadvantages of Ring Topology.

Answer: (Page 33)

Disadvantages of Ring Topology

¾ Unidirectional Traffic

9 A break in a ring I.e. a disabled station can disable the entire network

How are lost acknowledgment and a lost frame handled at the sender site? [5]

Answer:

A lost or damaged frame is handled in the same way by the receiver; when the receiver receives a damaged frame, it discards it, which essentially means the frame is lost. The receiver remains silent about a lost frame and keeps its value of R.

Question No: 40 ( Marks: 5 )

Explain Protocol Data Unit (PDU)?

Answer: (Page 221)

Protocol Data Unit (PDU)

The data unit in the LLC level is called the Protocol Data unit (PDU) The PDU contains 4 fields familiar from HDLC:

-A destination service access point (DSAP)
-A source service access point (SSAP)
-A control field